I wish to apply a sequence of functions to an object (each of the functions may return the same or modified object) and get the ultimate result returned by the last function.
Is there an idiomatic Scala way to turn this (pseudocode):
val pipeline = ListMap(("a" -> obj1), ("b" -> obj2), ("c" -> obj3))
into this?
val initial_value = Something("foo", "bar")
val result = obj3.func(obj2.func(obj1.func(initial_value)))
The pipeline is initialized at runtime and contains an undetermined number of "manglers".
I tried with foreach but it requires an intermediate var to store the result, and foldLeft only works on types of ListMap, while the initial value and the result are of type Something.
Thanks
This should do it:
pipeline.foldLeft(initial_value){case (acc, (k,obj)) => obj.func(acc)}
No idea why pipeline contains pairs, though.
Assuming input and output types are the same, I'd go with a reduceLeft and composition by andThen:
def pipe[A](a: A, funcs: List[A => A]): A = funcs.reduceLeft(_ andThen _)(a)
I think foldLeft is the right choice:
val pipeline = List("a"-> func1, "b"-> func2, "c"-> func3)
...
val result = pipeline.foldLeft(initial_value) {case (acc,(key,func)) => func(acc)}
Get rid of your keys, first:
pipeline.values.foldLeft(initial_value)((a, f) => f.func(a))
Related
I am getting empty list when I am trying to create the list with :: operator. My code looks like this:
def getAllInfo(locks: List[String]): List[LockBundle] = DB.withTransaction { implicit s =>
val myList = List[LockBundle]()
locks.foreach(
l => findForLock(l) :: myList
)
myList
}
def findForLock(lock: String): Option[LockBundle] = { ... }
Any suggestion?
Use flatMap
locks.flatMap(l => findForLock(l))
Your code becomes
def getAllInfo(locks: List[String]): List[LockBundle] = DB.withTransaction { implicit s =>
locks.flatMap(l => findForLock(l))
}
Alternatively you could use map and flatten. Something like this locks.map(l => findForLock(l)).flatten
Functional programming is all about transformations. You just have to transform your existing list into another list using a transformation which is your function findForLock.
Problem with your code
val myList = List[LockBundle]()
locks.foreach(
l => findForLock(l) :: myList
)
myList
First of all foreach returns Unit so, you use foreach for side effecting operations and not transformations. As you need transformation so do not use foreach.
Next, findForLock(l) :: myList gives you a value but this gets ignored as there is no one who is storing the value generated. So, in order to store the value use accumulator and pass it as a function parameter in case of recursion.
Correcting your code
If you want to do in your way. You need to use the accumulator.
First fix your types findForLock(l) returns Option, You list is of type List[LockBundle] so change the list type to List[Option[LockBundle]].
In order to get List[LockBundle] from List[Option[LockBundle]] Just do flatten on List[Option[LockBundle]] list. See below code snippet
var myList = List[Option[LockBundle]]()
locks.foreach(
l => myList = findForLock(l) :: myList
)
myList.flatten
The above way is not functional and is not recommended.
Your code doesn't work, because foreach combinator calls given closure for each element, but all you do here is to return expression findForLock(l) :: myList which is discarded.
As pamu suggested, you can use flatMap on a function to map each element to values returned by findForLock and flatten that list, which turns Option into element of the list if it's Some or nothing if it's None.
Keep in mind that this works only because there is an implicit conversion from Option to Seq, in general flatMap works only if you return the same type as given monad (that in this case is List or Option).
in my main program i receive inputs like -
key1=value1 key2=value2
Now what I want is to create a map out of it. I know the imperative way of doing this where I would get Array[String] that can be foreach and then split by "=" and then key and value can be used to form a Map.
is there a good functional and readable way to achieve this?
Also It will be great if I can avoid mutable Map and I want to avoid initial Dummy value initialization.
def initialize(strings: Array[String]): Unit = {
val m = collection.mutable.Map("dummy" -> "dummyval")
strings.foreach(
s => {
val keyVal:Array[String] = s.split("=")
m += keyVal(0) -> keyVal(1)
})
println(m)
}
you can just use toMap().
However, converting from array to tuple is not quite trivial:
How to convert an Array to a Tuple?
scala> val ar = Array("key1=value1","key2=value2")
ar: Array[String] = Array(key1=value1, key2=value2)
scala> ar.collect(_.split("=") match { case Array(x,y) => (x,y)}).toMap
res10: scala.collection.immutable.Map[String,String] = Map(key1 -> value1, key2 -> value2)
Maybe you have to call Function.unlift for intellij
val r = ar.collect(Function.unlift(_.split("=") match { case Array(x, y) => Some(x, y)})).toMap
similar to above but using only 'map'
ar.map(_.split("=")).map(a=>(a(0), a(1))).toMap
You can use Scopt to do the command line argument parsing in a neat way.
I have following list
val a = List(("name1","add1","city1",10),("name1","add1","city1",10),
("name2","add2","city2",10),("name2","add2","city2",20),("name3","add3","city3",20))
I want distinct element from above list based on first three values of tuple. Fourth value should not be consider while finding distinct elements from list.
I want following output:
val output = List(("name1","add1","city1",10),("name2","add2","city2",10),
("name3","add3","city3",20))
Is it possible to get above output?
As per my knowledge, distinct works if whole tuple/value is duplicated. I tried out with distinct like following:
val b = List(("name1","add1","city1",10),("name1","add1","city1",10),("name2","add2","city2",10),
("name2","add2","city2",20),("name3","add3","city3",20)).distinct
but it gives output as -
List(("name1","add1","city1",10),("name2","add2","city2",10),
("name2","add2","city2",20),("name3","add3","city3",20))
Any alternate approach will also appreciated.
Use groupBy like this
a.groupBy( v => (v._1,v._2,v._3)).keys.toList
This constructs a Map where each key is by definition a unique triplet as required in the lambda function above.
Should it include also the last element in the tuple, fetch the first element for each key, like this
a.groupBy( v => (v._1,v._2,v._3)).mapValues(_.head)
If the order of the output list isn't important (i.e. you are happy to get List(("name3","add3","city3",20),("name1","add1","city1",10),("name2","add2","city2",10))), the following works as specified:
a.groupBy(v => (v._1,v._2,v._3)).values.map(_.head).toList
(Due to Scala collections design, you'll see the order kept for output lists up to 4 elements, but above that size HashMap will be used.) If you do need to keep the order, you can do something like (generalizing a bit)
def distinctBy[A, B](xs: Seq[A], f: A => B) = {
val seen = LinkedHashMap.empty[B, A]
xs.foreach { x =>
val key = f(x)
if (!seen.contains(key)) { seen.update(key, x) }
}
seen.values.toList
}
distinctBy(a, v => (v._1, v._2, v._3))
You could try
a.map{case x#(name, add, city, _) => (name,add,city) -> x}.toMap.values.toList
To make sure you have the first one in list kept,
type String3 = (String, String, String)
type String3Int = (String, String, String, Int)
a.foldLeft(collection.immutable.ListMap.empty[String3, String3Int]) {
case (a, b) => if (a.contains((b._1, b._2, b._3))) {
a
} else a + ((b._1, b._2, b._3) -> b)
}.values.toList
On simple solution would be to convert the List to a Set. Sets don't contain duplicates: check the documentation.
val setOfTuples = a.toSet
println(setOfTuples)
Output: Set((1,1), (1,2), (1,3), (2,1))
I need to group list of tuples in some unique way.
For example, if I have
val l = List((1,2,3),(4,2,5),(2,3,3),(10,3,2))
Then I should group the list with second value and map with the set of first value
So the result should be
Map(2 -> Set(1,4), 3 -> Set(2,10))
By so far, I came up with this
l groupBy { p => p._2 } mapValues { v => (v map { vv => vv._1 }).toSet }
This works, but I believe there should be a much more efficient way...
This is similar to this question. Basically, as #serejja said, your approach is correct and also the most concise one. You could use collection.breakOut as builder factory argument to the last map and thereby save the additional iteration to get the Set type:
l.groupBy(_._2).mapValues(_.map(_._1)(collection.breakOut): Set[Int])
You shouldn't probably go beyond this, unless you really need to squeeze the performance.
Otherwise, this is how a general toMultiMap function could look like which allows you to control the values collection type:
import collection.generic.CanBuildFrom
import collection.mutable
def toMultiMap[A, K, V, Values](xs: TraversableOnce[A])
(key: A => K)(value: A => V)
(implicit cbfv: CanBuildFrom[Nothing, V, Values]): Map[K, Values] = {
val b = mutable.Map.empty[K, mutable.Builder[V, Values]]
xs.foreach { elem =>
b.getOrElseUpdate(key(elem), cbfv()) += value(elem)
}
b.map { case (k, vb) => (k, vb.result()) } (collection.breakOut)
}
What it does is, it uses a mutable Map during building stage, and values gathered in a mutable Builder first (the builder is provided by the CanBuildFrom instance). After the iteration over all input elements has completed, that mutable map of builder values is converted into an immutable map of the values collection type (again using the collection.breakOut trick to get the desired output collection straight away).
Ex:
val l = List((1,2,3),(4,2,5),(2,3,3),(10,3,2))
val v = toMultiMap(l)(_._2)(_._1) // uses Vector for values
val s: Map[Int, Set[Int] = toMultiMap(l)(_._2)(_._1) // uses Set for values
So your annotated result type directs the type inference of the values type. If you do not annotate the result, Scala will pick Vector as default collection type.
I have a collection which I want to map to a new collection, however each resulting value is dependent on the value before it in some way.I could solve this with a leftFold
val result:List[B] = (myList:List[A]).foldLeft(C -> List.empty[B]){
case ((c, list), a) =>
..some function returning something like..
C -> (B :: list)
}
The problem here is I need to iterate through the entire list to retrieve the resultant list. Say I wanted a function that maps TraversableOnce[A] to TraversableOnce[B] and only evaluate members as I call them?
It seems to me to be a fairly conventional problem so Im wondering if there is a common approach to this. What I currently have is:
implicit class TraversableOnceEx[T](val self : TraversableOnce[T]) extends AnyVal {
def foldyMappyFunction[A, U](a:A)(func:(A,T) => (A,U)):TraversableOnce[U] = {
var currentA = a
self.map { t =>
val result = func(currentA, t)
currentA = result._1
result._2
}
}
}
As far as functional purity goes, you couldn't run it in parallel, but otherwise it seems sound.
An example would be;
Return me each element and if it is the first time that element has appeared before.
val elements:TraversableOnce[E]
val result = elements.mappyFoldyFunction(Set.empty[E]) {
(s, e) => (s + e) -> (e -> s.contains(e))
}
result:TraversableOnce[(E,Boolean)]
You might be able to make use of the State Monad. Here is your example re-written using scalaz:
import scalaz._, Scalaz._
def foldyMappy(i: Int) = State[Set[Int], (Int, Boolean)](s => (s + i, (i, s contains(i))))
val r = List(1, 2, 3, 3, 6).traverseS(foldyMappy)(Set.empty[Int])._2
//List((1,false), (2,false), (3,false), (3,true), (6,false))
println(r)
It is look like you need SeqView. Use view or view(from: Int, until: Int) methods for create a non-strict view of list.
I really don't understand your example as your contains check will always result to false.
foldLeft is different. It will result in a single value by aggregating all elements of the list.
You clearly need map (List => List).
Anyway, answering your question about laziness:
you should use Stream instead of List. Stream doesn't evaluate the tail before actually calling it.
Stream API